The expression ((1+i)n/(1-i)n-2) equals

Question

The expression ((1+i)n/(1-i)n-2) equals (A) -in+1 (B) in+1 (C) -2in+1 (D) 1

Tardigrade Admin · Accepted Answer

((1+i)n/(1-i)n-2)=((1+i)n/(1-i)n(1-i)-2) =((1+i/1-i))n(1-i)2 =((1+i/1-i) × (1+i/1+i))n(1+i2-2 i) =((1+2 i+i2/1-i2))n(1+i2-2 i) =((1+2 i-1/1-(-1)))n(1-1-2 i) [∵ i2=-1] =((2 i/2))n(-2 i) =in(-2 i)=-2 in+1

Q. The expression $\frac{( 1 + i ) ^{n}}{( 1 - i ) ^{n - 2}}$ equals

$- i^{n + 1}$

$i^{n + 1}$

$- 2 i^{n + 1}$

$1$

Q. The expression (1−i)n−2(1+i)n​ equals

−in+1

in+1

−2in+1

1

(1−i)n−2(1+i)n​=(1−i)n(1−i)−2(1+i)n​ =(1−i1+i​)n(1−i)2 =(1−i1+i​×1+i1+i​)n(1+i2−2i) =(1−i21+2i+i2​)n(1+i2−2i) =(1−(−1)1+2i−1​)n(1−1−2i) [∵i2=−1] =(22i​)n(−2i) =in(−2i)=−2in+1

Q. The expression $\frac{( 1 + i ) ^{n}}{( 1 - i ) ^{n - 2}}$ equals

$- i^{n + 1}$

$i^{n + 1}$

$- 2 i^{n + 1}$

$1$